Regex string between square brackets only if '.' is within string-CodePudding

I'm trying to detect the text between two square brackets in Python however I only want the result where there is a "." within it.

I currently have [(.*?] as my regex, using the following example:

String To Search: CASE[Data Source].[Week] = 'THIS WEEK'

Result: Data Source, Week

However I need the whole string as [Data Source].[Week], (square brackets included, only if there is a '.' in the middle of the string). There could also be multiple instances where it matches.

CodePudding user response：

You might write a pattern matching [...] and then repeat 1 or more times a . and again [...]

\[[^][]*](?:\.\[[^][]*])

Explanation

\[[^][]*] Match from [...] using a negated character class
(?: Non capture group to repeat as a whole part
- \.\[[^][]*] Match a dot and again [...]
) Close the non capture group and repeat 1 times

See a regex demo.

To get multiple matches, you can use re.findall

import re

pattern = r"\[[^][]*](?:\.\[[^][]*]) "

s = ("CASE[Data Source].[Week] = 'THIS WEEK'\n"
            "CASE[Data Source].[Week] = 'THIS WEEK'")

print(re.findall(pattern, s))

Output

['[Data Source].[Week]', '[Data Source].[Week]']

If you also want the values of between square brackets when there is not dot, you can use an alternation with lookaround assertions:

\[[^][]*](?:\.\[[^][]*]) |(?<=\[)[^][]*(?=])

Explanation

\[[^][]*](?:\.\[[^][]*]) The same as the previous pattern
| Or
(?<=\[)[^][]*(?=]) Match [...] asserting [ to the left and ] to the right

See another regex demo

CodePudding user response：

I think an alternative approach could be:

import re

pattern = re.compile("(\[[^\]]*\]\.\[[^\]]*\])")

print(pattern.findall(sss))

OUTPUT

['[Data Source].[Week]']